Ok, I did some math with lots of simplifications to make it easier to calculate.

Let's assume that we're interested in only 90 degrees of your camera's field of view. As it's facing straight up into the sky, any central vertical cross-section of this "cone of view" is a 45-90-45 triangle, which we can further divide in half through camera's central axis to get two smaller, but similar

isosceles right triangles.

I hope you can see where I'm going with all this, because now the math gets really simple:

- assumption #1: object's speed was 10 m/s = 36kmph = 22.5 mph

- assumption #2: it flew over camera's FOV (reduced to 90 deg) in 10 seconds

This means that it took 5 seconds for the object to fly over our smaller triangle's top catete, which makes it 10m/s x 5 sec = 50 meters (~165ft) long, and also 50 meters high (because both catetes are of the same length). If this was true for the object in the composite image from my previous post, it would mean there's 10 meter distance between each consecutive frame and the object is only a few meters in diameter.

Basically, what I'm saying here is that it would be possible for an object ~160ft high carried by the ~20mph wind to get in and out of your camera's FOV in approximately 20 seconds - but only if its direction agrees with the general wind direction at the time. As I know neither your location nor the directions in your camera's images, I can't tell if this is the case - so it's your call.

P.S. - there's an interesting physical limit for this kind of calculations: any object higher than 187500 miles traveling through the 90 degree FOV in less than 2 seconds would have to be moving faster than the speed of light.